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I have a question regarding the number of water molecules produced by cellular respiration. At the level of the electron transport system (ETS), the equation regarding O2 as the final electron acceptor is: 1/2O2 + 2é + 2H+ --- H20. Considering that the 10 NADH molecules entering the ETS will contribute to 10 é pairs (and hence to the consumption of 10/2 = 5 O2 molecules), and that the 2 FADH2 molecules entering the ETS will contribute to 2 é pairs (and use 2/2 = 1 O2 molecule). Then what about water? According to this equation, a total of 12 molecules of water would be produced. How should we relate this to the general equation of glucose breakdown: glucose + 6 O2 ---- 6 CO2 + 6 H20 + E ??? This should mean that 6 water molecules are consumed at some other point to account for the global equation??
Any help would be appreciated, thanks.
Last edited by biofan on Fri Mar 17, 2006 8:19 pm, edited 1 time in total.
Sorry if my question lacked in clarity. I wasn't interested so much about the idea of water production per se, but rather about the understanding of the equations (trying to balance the general equation that states that 6 molecules of water will be formed for each glucose molecule broken down, with the equation that accounts for the oxygen consumption (and water production) at the end of the ETS. This equation states that for 1/2 molecule of O2 consumed by accepting an electron pair, there will be 1 molecule of water formed. Hence, considering that the 10 NADH molecules entering the ETS will contribute 10 electron pairs to the ETS, and the 2 FADH2 molecules contributes to 2 electron pairs; 12 molecules of water will be formed for each glucose molecule being broken down... I then presume that somewhere upstream in the pathways (glycolysis? transition reaction? Krebs cycle?), there should be 6 molecules of water being consumed, but where?
Well, i can tell you this:
Glycolysis also produces one molecule of water from 2-phosphoglycerate to phosphoenolpiruvate.(so 2 extra molecules of water are produced this way). On the other hand the Krebs cycle uses up 2 molecules of water per glucose when it turns fumarate into malate.
The answer to your question lies at the electron transport chain. i have been killing my brains trying to figure out how. I know 1000% that for every 2 NADH or FADH2 molecules one water molecule is produced. What i can not understand is how the heck?!? Cause every NADH molecule carries 2 electrons, and it should be enough to reduce one water molecule. Sorry, i just never realised this problem before. I will post it here as soon as i have an answer
"As a biologist, I firmly believe that when you're dead, you're dead. Except for what you live behind in history. That's the only afterlife" - J. Craig Venter
Thanks Mr Mistery. It is one of those questions that seems so obvious at first sight that you just don't pay attention to it... just like the majority of textbooks on the subject I've consulted thus far! I don't know if this is a similar phenomenon to the one in the journalistic circle, where it sometimes seems like everybody is copying each other, but it seems like no one seemed to have noticed that things didn't balance out in terms of water production... Now I don't intend to say that things don't happen this way (as I said; there should be an overall consumption of 6 molecules of water somewhere upstream of the ETS to account for it), but rather point out the fact that the most common textbooks on the subject should have mentionned it. Hoping someone can help me figure out this one... Thanks in advance!
krebs cycle uses H2O? Where? Maybe it says that 3 water molecules take place in the Krebs cycle. Cause conversion of citrate to isocitrate is done by extracting and then inserting a water molecule.
Colleagues: Well I think I have found another point of water consumption. Most diagrams of the Kreb's Cycle in text books show water being released as a waste product between Citrate and cis-Aconitate and then this is usually followed by an illustration of a single water molecule being consumed in converting cis Aconitate into D-isocitrate. Net water gain zero. One in one out. Later in the Kreb's cycle a single water molecule is consumed in converting fumarate into malate. This gives you one water molecule consumed per acetyl Co A or two consumed per glucose catabolized just within the Kreb's. These 2 Kreb cycle consumed waters balance against the 2 produced via enolase during the glycolytic conversion of two 2-phosphoglycerate molecules into two phospho-enol pyruvate molecules. But what most text books never show is that during the Kreb's cycle and specifically during the combination reaction of acetylCoA plus oxaloacteate to form citrate via citrate synthase that a single water molecule is consumed. That is one additional water consumption per acetyl CoA then typically illustrated or 2 water consumptions per each catabolized glucose during the Kreb's, So if 10 NADH's and 2 FADH2s each drop off 2e-to the ETC with each combining with 1/2O2 and 2H+ to reduce the oxygen then 12 water molecules should be formed. If we subtract the 2 waters consumed during citrate formation that brings us down to 10 water molecules net. Which is still 4 more then the net equation suggests (C6H12O6 +6 O2 --> 6CO2 +6H2O). So somewhere else 4 water molecules must be consumed...but where is still a mystery. Maybe in the oxidation reactions of pyruvate that from acetyl CoA or maybe hidden within the reactions of glycolysis, The website: https://www.boundless.com/image/the-citric-acid-cycle/ shows the water productions and consumptions I have mentioned above. The key may lie in searching for and finding better metabolic pathway diagrams that dont conceal these water consumptions similar to the way that most textbooks never ever show that citrate formation consumes water as acetyl CoA and oxaloacetate combine. The other issue is that I have assumed that only one water is consumed in converting fumarate to malate or in forming citrate...what if 2 or more are consumed per each of these conversions. I think we are closer to a resolution but we still have some more research to do to get the bottom of this issue.
I did a google search for "aldolase glycolysis" and found a biochem website at Georgia Tech that posted a powerpoint link
In that powerpoint they show that the aldolase enzyme executing a reverse aldol condensation that consumes one water molecule per Dihydroxyacetone produced during the splitting of one fructose 1,6 bis phosphate. So with the 2 consumed during citrate formation we are now down to 12 -3 waters or 9 waters produced. 3 are still in excess of the equation but I am sure that those can be found by looking at better biochemistry resources where they show more details about the individual reactions occurring in the active sites of the enzymes.
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